A 70 kg bungee jumper steps off a bridgt with a light bungee cord tied to her bo

Question

A 70 kg bungee jumper steps off a bridgt with a light bungee cord tied to her body and to the bridge. The unstretched length of the cord is 13.5 m. The jumper reaches reaches the bottom of her motion 36.5 m below the bridge before bouncing back we find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 13.5 m free fall and a 23.0 m section of simple harmonic Oscillation. (Use the exact values you enter in answer(s) to make later calculation(s).) For the free-fall part, what is the appropriate analysis model to describe her motion. particle under constant acceleration particle under constant angular acceleration particle in simple harmonic motion For what time interval is she in free fall? s For the the simple harmonic oscillation part of the plunge, is the system of the bungee jumper, the spring, and the Earth isolated or non-isolated? isolated non-isolated From your response in part (c) find the spring constant of the bungee cord. N/m What is the location of the equilibrium point where the spring force balances the gravitational force exerted on the jumper? m below the bridge What is the angular frequency of the oscillation? rad/s What time interval is required for the cord to stretch by 23.0 m? s What is the total time interval for the entire 36.5 m drop? s

Explanation / Answer

b) tf = sqrt [2 L / g]

= sqrt [2 * 13.5 / 9.8]

= 1.66 s

d) k = 2 m g ht / (ht - L)2 = (2 * 71 * 9.8 * 36.5) / (36.5 - 13.5)2

= 96.02 N/m

e) delta Leq = m g / k = (71 * 9.8) / 96.02

= 7.25 m

f) w = sqrt [k/m] = sqrt [96.02 / 71]

= 1.16 rad/s

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